EquationsMCQPYQ Nov 18Question 1052 of 221
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Let α\displaystyle \alpha and β\displaystyle \beta be the roots of x2+7x+12=0\displaystyle x^2 + 7x + 12 = 0. Then the value of (αβ+βα)\displaystyle (\frac{\alpha}{\beta} + \frac{\beta}{\alpha}) will be:

Options

A712+127\displaystyle \frac{7}{12} + \frac{12}{7}
B49144+14449\displaystyle \frac{49}{144} + \frac{144}{49}
C−9112\displaystyle -\frac{91}{12}
DNone of these
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Correct Answer

✅ Option d — None of these

All Options:

  • A712+127\displaystyle \frac{7}{12} + \frac{12}{7}
  • B49144+14449\displaystyle \frac{49}{144} + \frac{144}{49}
  • C−9112\displaystyle -\frac{91}{12}
  • DNone of these

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Detailed Solution & Explanation

Given the quadratic equation:
x2+7x+12=0x^2 + 7x + 12 = 0
Comparing with ax2+bx+c=0\displaystyle ax^2 + bx + c = 0, we have:
a=1,b=7,c=12\displaystyle a = 1, b = 7, c = 12

According to Vieta's formulas:
Sum of roots: α+β=−ba=−7\displaystyle \alpha + \beta = -\frac{b}{a} = -7
Product of roots: αβ=ca=12\displaystyle \alpha\beta = \frac{c}{a} = 12

Now, we simplify the required expression:
αβ+βα=α2+β2αβ\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}
Using the algebraic identity α2+β2=(α+β)2−2αβ\displaystyle \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta:
α2+β2=(−7)2−2(12)=49−24=25\alpha^2 + \beta^2 = (-7)^2 - 2(12) = 49 - 24 = 25
Now substitute this back:
αβ+βα=2512\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{25}{12}

Alternatively, by factoring the given quadratic equation:
x2+7x+12=(x+3)(x+4)=0  ⟹  α=−3,β=−4x^2 + 7x + 12 = (x+3)(x+4) = 0 \implies \alpha = -3, \beta = -4
Substituting these values:
αβ+βα=−3−4+−4−3=34+43=9+1612=2512\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{-3}{-4} + \frac{-4}{-3} = \frac{3}{4} + \frac{4}{3} = \frac{9 + 16}{12} = \frac{25}{12}
Since 2512\displaystyle \frac{25}{12} is not present in Option a, b, or c, the correct choice is "None of these".
**Option d**

About This Chapter: Equations

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Linear, Quadratic and Cubic Equations

This chapter covers Linear, Quadratic and Cubic Equations and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

More Questions from Equations

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In a multiple choice question paper consisting of 100\displaystyle 100 questions of 1\displaystyle 1 mark each, a candidate gets 60%\displaystyle 60\% marks. If the candidate attempted all questions and there was a penalty of 0.25\displaystyle 0.25 marks for wrong answers is:

The values of x\displaystyle x and y\displaystyle y satisfying the equations 3x+y+2x−y=3\displaystyle \frac{3}{x+y} + \frac{2}{x-y} = 3 and 2x+y+3x−y=233\displaystyle \frac{2}{x+y} + \frac{3}{x-y} = \frac{23}{3} given by

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